# Smooth vector fields on a surface modulo diffeomorphisms

Let $$\Sigma$$ be a two-dimensional connected smooth manifold without boundary. (Feel free to assume it is compact and orientable.)

Let $$\mathcal{X}(\Sigma)$$ denote the smooth vector fields on $$\Sigma$$ and let $$\operatorname{Diff}(\Sigma)$$ denote the group of diffeomorphisms of $$\Sigma$$. Clearly $$\operatorname{Diff}(\Sigma)$$ acts on $$\mathcal{X}(\Sigma)$$.

I would expect naively that the space of orbits $$\mathcal{M}:=\mathcal{X}(\Sigma)/\operatorname{Diff}(\Sigma)$$ would be finite-dimensional. Is this actually the case?

Question

What can one say about $$\mathcal{M}$$ in general?

This is not finite dimensional. For example, consider non-vanishing vector fields on $$T^2=\mathbb R^2/\mathbb Z^2$$, transversal to vertical circles. Any such field defines a self diffeo $$S^1\to S^1$$ on a vertical circle, called the return map. Suppose that such a diffeo $$\varphi$$ has $$n$$ fixed points $$x_i$$ (they correspond to closed orbits of the field). Then for each fixed point $$x_i\in S^1$$ we have a linear map on the tangent space $$d\varphi: T_{x_i}S^1\to T_{x_i}S^1$$ given by multiplication by $$\alpha_i\in \mathbb R^*$$. Such $$\alpha_i$$ is an invariant of a vector field under diffeomorphisms. And since the number of closed orbits can be arbitrary and these $$\alpha_i$$'s are independent, we see that the dimension is not finite.
If you want something which is finite-dimensional, one can restrict to area-preserving vector fields which are same thing as closed $$1$$-forms. Now the spaces of minimal closed $$1$$-forms on a closed genus $$g$$ surface is indeed finite-dimensional by a theorem of Calabi. Each such form is the real part of a holomorphic $$1$$-form for a certain complex structure on the surface. (the reference is: E. Calabi, An intrinsic characterization of harmonic 1-forms, Global Analysis, Papers in Honor of K. Kodaira, 1969)